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derivate (x)/(((x^2)+4)^2)

 Dec 16, 2015

Best Answer 

 #2
avatar+26400 
+15

derivate (x)/(((x^2)+4)^2)

 

\(\begin{array}{lcl} y = \frac{x}{(4+x^2)^2} &=& \frac{x}{(4+x^2)\cdot (4+x^2)} \end{array} \)

 

 

Use the general quotient rule:

\(\begin{array}{lcll} y & = & \frac{f(x)}{ g(x)\cdot h(x) }\\ y' &=& y\cdot \left[ \frac{ f'(x) } {f(x)} - \frac{ g'(x) }{g(x)} -\frac{ h'(x) }{h(x)} \right] \\ y' &=& \frac{f(x)}{ g(x)\cdot h(x) }\cdot \left[ \frac{ f'(x) } {f(x)} - \frac{ g'(x) }{g(x)} -\frac{ h'(x) }{h(x)} \right] \\ y' &=& \frac{ f'(x) } { g(x)\cdot h(x)} - \frac{ f(x)\cdot g'(x) }{g^2(x)\cdot h(x)} -\frac{ f(x)\cdot h'(x) }{g(x)\cdot h^2(x)} \\ y' &=& \frac{ f'(x)\cdot g(x)\cdot h(x) - f(x)\cdot g'(x)\cdot h(x) - f(x)\cdot h'(x) \cdot g(x) } {g^2(x)\cdot h^2(x)} \\\\ \end{array}\\ \begin{array}{lcll} f(x) &=& x & f'(x) = 1 \\ g(x) &=& 4+x^2 & g'(x) = 2x \\ h(x) &=& 4+x^2 & h'(x) = 2x \\\\ \end{array}\\ \begin{array}{lcll} y' &=& \frac{ 1\cdot (4+x^2)\cdot (4+x^2) - x\cdot 2x\cdot (4+x^2) - x\cdot 2x \cdot (4+x^2) } {(4+x^2)^2\cdot (4+x^2)^2} \\ y' &=& \frac{ (4+x^2)^2 - 2\cdot [~ x\cdot 2x\cdot (4+x^2)~] } {(4+x^2)^4} \\ y' &=& \frac{ (4+x^2)^2 - 4\cdot x^2\cdot (4+x^2) } {(4+x^2)^4} \\ y' &=& \frac{ (4+x^2)^2 }{(4+x^2)^4} - \frac{ 4\cdot x^2\cdot (4+x^2) } {(4+x^2)^4} \\ y' &=& \frac{ 1 }{(4+x^2)^2} - \frac{ 4\cdot x^2 } {(4+x^2)^3} \\ \end{array}\)

 

laugh

 Dec 16, 2015
 #1
avatar
+5

Possible derivation:
d/dx(x/(x^2+4)^2)
Use the product rule, d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = x and v = 1/(x^2+4)^2:
  =  (d/dx(x))/(4+x^2)^2+x (d/dx(1/(4+x^2)^2))
The derivative of x is 1:
  =  x (d/dx(1/(4+x^2)^2))+1/(x^2+4)^2
Using the chain rule, d/dx(1/(x^2+4)^2) =  d/( du)1/u^2 ( du)/( dx), where u = x^2+4 and ( d)/( du)(1/u^2) = -2/u^3:
  =  1/(4+x^2)^2+(-2 d/dx(4+x^2))/(x^2+4)^3 x
Differentiate the sum term by term:
  =  1/(4+x^2)^2-(d/dx(4)+d/dx(x^2) 2 x)/(4+x^2)^3
The derivative of 4 is zero:
  =  1/(4+x^2)^2-(2 x (d/dx(x^2)+0))/(4+x^2)^3
Simplify the expression:
  =  1/(4+x^2)^2-(2 x (d/dx(x^2)))/(4+x^2)^3
Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:
  =  1/(4+x^2)^2-(2 x 2 x)/(4+x^2)^3
Simplify the expression:
Answer: | =  -(4 x^2)/(4+x^2)^3+1/(4+x^2)^2
 

 Dec 16, 2015
 #2
avatar+26400 
+15
Best Answer

derivate (x)/(((x^2)+4)^2)

 

\(\begin{array}{lcl} y = \frac{x}{(4+x^2)^2} &=& \frac{x}{(4+x^2)\cdot (4+x^2)} \end{array} \)

 

 

Use the general quotient rule:

\(\begin{array}{lcll} y & = & \frac{f(x)}{ g(x)\cdot h(x) }\\ y' &=& y\cdot \left[ \frac{ f'(x) } {f(x)} - \frac{ g'(x) }{g(x)} -\frac{ h'(x) }{h(x)} \right] \\ y' &=& \frac{f(x)}{ g(x)\cdot h(x) }\cdot \left[ \frac{ f'(x) } {f(x)} - \frac{ g'(x) }{g(x)} -\frac{ h'(x) }{h(x)} \right] \\ y' &=& \frac{ f'(x) } { g(x)\cdot h(x)} - \frac{ f(x)\cdot g'(x) }{g^2(x)\cdot h(x)} -\frac{ f(x)\cdot h'(x) }{g(x)\cdot h^2(x)} \\ y' &=& \frac{ f'(x)\cdot g(x)\cdot h(x) - f(x)\cdot g'(x)\cdot h(x) - f(x)\cdot h'(x) \cdot g(x) } {g^2(x)\cdot h^2(x)} \\\\ \end{array}\\ \begin{array}{lcll} f(x) &=& x & f'(x) = 1 \\ g(x) &=& 4+x^2 & g'(x) = 2x \\ h(x) &=& 4+x^2 & h'(x) = 2x \\\\ \end{array}\\ \begin{array}{lcll} y' &=& \frac{ 1\cdot (4+x^2)\cdot (4+x^2) - x\cdot 2x\cdot (4+x^2) - x\cdot 2x \cdot (4+x^2) } {(4+x^2)^2\cdot (4+x^2)^2} \\ y' &=& \frac{ (4+x^2)^2 - 2\cdot [~ x\cdot 2x\cdot (4+x^2)~] } {(4+x^2)^4} \\ y' &=& \frac{ (4+x^2)^2 - 4\cdot x^2\cdot (4+x^2) } {(4+x^2)^4} \\ y' &=& \frac{ (4+x^2)^2 }{(4+x^2)^4} - \frac{ 4\cdot x^2\cdot (4+x^2) } {(4+x^2)^4} \\ y' &=& \frac{ 1 }{(4+x^2)^2} - \frac{ 4\cdot x^2 } {(4+x^2)^3} \\ \end{array}\)

 

laugh

heureka Dec 16, 2015

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